$\int_0^3 \sqrt{9 - x^2} dx = $

Using the standard integral formula:

$\int \sqrt{a^2 - x^2} dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\left(\dfrac{x}{a}\right)$

Applying the limits from $0$ to $3$ with $a = 3$:

$\int_0^3 \sqrt{9 - x^2} dx = \left[ \dfrac{x}{2}\sqrt{9 - x^2} + \dfrac{9}{2}\sin^{-1}\left(\dfrac{x}{3}\right) \right]_0^3$

$= \left( 0 + \dfrac{9}{2}\sin^{-1}(1) \right) - \left( 0 + \dfrac{9}{2}\sin^{-1}(0) \right)$

$= \dfrac{9}{2} \times \dfrac{\pi}{2} - 0$

$= \dfrac{9\pi}{4}$

Alternatively, the integral represents the area of a quarter circle of radius $3$ in the first quadrant, which is $\dfrac{1}{4} \times \pi(3)^2 = \dfrac{9\pi}{4}$.